Use the Laplace transform to solve the given system of differential equations.(d2x)(dt2)+(d2y)(dt2)=t2 (d2x)(dt2)−(d2y)(dt2)=4t x(0)=5,x′(0)=0,y(0)=0,y′(0)=0

Name: Use the Laplace transform to solve the given system of differential equations.(d2x)(dt2)+(d2y)(dt2)=t2 (d2x)(dt2)−(d2y)(dt2)=4t x(0)=5,x′(0)=0,y(0)=0,y′(0)=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: Use the Laplace transform to solve the given system of differential equations.(d2x)(dt2)+(d2y)(dt2)=t2 (d2x)(dt2)−(d2y)(dt2)=4t x(0)=5,x′(0)=0,y(0)=0,y′(0)=0

Step 1Given that(d2x)(dt2)+(d2y)(dt2)=t2…(A)(d2x)(dt2)−(d2y)(dt2)=4t…(B)x(0)=5,x′(0)=0,y(0)=0,y′(0)=0adding equation( A) and (B)⇒2(d2x)(dt2)=(t2+4t)(d2x)(dt2)=(t2+4t)2(d2x)(dt2)=(t2)2+2tStep 2taking Laplace transform on both sides,s2X(s)−sx(0)−x′(0)=12+2s2+2s2s2X(s)−5s=1(s3)+2s2s2X(s)=(1+2s+5s4)s5X(s)=1s5+2s4+5staking Inverse Laplace transformx(t)=t44!+(2t3)3!+5u(t)Step 3u(t)=1,t≥0=0,t&lt;0∴x(t)=(t4)24+(t3)3+5 for t≥0And Subtracting equation (B) from (A)⇒2(d2y)(dt2)=t2−4t(d2y)(dt2)=(t2−4t)2=(t2)2−2tTaking ILTs2Y(s)−sy(0)−y′(0)=122s3−2s2s2Y(s)=1(s3)−2(s2)=(1−2s)s3Y(s)=(1−2s)s5Step 4Y(s)=1s5−2s(s5)Y(s)=1s5−2(s4)Taking Inverse Laplace Transformy(t)=(t4)4!−(2t3)3!=(t4)24−(t3)3

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allhvasstH

Answered question

2021-01-25

Use the Laplace transform to solve the given system of differential equations.

\frac{(d^{2} x)}{(d t^{2})} + \frac{(d^{2} y)}{(d t^{2})} = \frac{t}{2}

\frac{(d^{2} x)}{(d t^{2})} - \frac{(d^{2} y)}{(d t^{2})} = 4 t

x (0) = 5, x^{'} (0) = 0,

y (0) = 0, y^{'} (0) = 0

Answer & Explanation

Daphne Broadhurst

Skilled2021-01-26Added 109 answers

Step 1
Given that
$\frac{(d^{2} x)}{(d t^{2})} + \frac{(d^{2} y)}{(d t^{2})} = \frac{t}{2} \dots (A)$
$\frac{(d^{2} x)}{(d t^{2})} - \frac{(d^{2} y)}{(d t^{2})} = 4 t \dots (B)$
$x (0) = 5, x^{'} (0) = 0,$
$y (0) = 0, y^{'} (0) = 0$
adding equation( A) and (B)
$\Rightarrow 2 \frac{(d^{2} x)}{(d t^{2})} = (t^{2} + 4 t)$
$\frac{(d^{2} x)}{(d t^{2})} = \frac{(t^{2} + 4 t)}{2}$
$\frac{(d^{2} x)}{(d t^{2})} = \frac{(t^{2})}{2} + 2 t$
Step 2
taking Laplace transform on both sides,
$s 2 X (s) - s x (0) - x^{'} (0) = \frac{1}{2} + \frac{2}{s^{2}} + \frac{2}{s^{2}}$
$s 2 X (s) - 5 s = \frac{1}{(s^{3})} + \frac{2}{s^{2}}$
$s 2 X (s) = \frac{(1 + 2 s + 5 s^{4})}{s^{5}}$
$X (s) = \frac{1}{s^{5}} + \frac{2}{s^{4}} + \frac{5}{s}$
taking Inverse Laplace transform
$x (t) = \frac{t^{4}}{4!} + \frac{(2 t^{3})}{3!} + 5 u (t)$
Step 3
$u (t) = 1, t \geq 0$
$= 0, t < 0$
$∴ x (t) = \frac{(t^{4})}{24} + \frac{(t^{3})}{3} + 5 for t \geq 0$
And Subtracting equation (B) from (A)
$\Rightarrow 2 \frac{(d^{2} y)}{(d t^{2})} = t^{2} - 4 t$
$\frac{(d^{2} y)}{(d t^{2})} = \frac{(t^{2} - 4 t)}{2} = \frac{(t^{2})}{2} - 2 t$
Taking ILT
$s^{2} Y (s) - s y (0) - y^{'} (0) = \frac{1}{2} \frac{2}{s^{3}} - \frac{2}{s^{2}}$
$s^{2} Y (s) = \frac{1}{(s^{3})} - \frac{2}{(s^{2})} = \frac{(1 - 2 s)}{s^{3}}$
$Y (s) = \frac{(1 - 2 s)}{s^{5}}$
Step 4
$Y (s) = \frac{1}{s^{5}} - \frac{2 s}{(s^{5})}$
$Y (s) = \frac{1}{s^{5}} - \frac{2}{(s^{4})}$
Taking Inverse Laplace Transform
$y (t) = \frac{(t^{4})}{4!} - \frac{(2 t^{3})}{3!}$
$= \frac{(t^{4})}{24} - \frac{(t^{3})}{3}$

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